Respuesta :
Answer:
The estimated standard error for the sampling distribution of differences in sample proportions is 0.0382.
Step-by-step explanation:
To solve this question, we need to understand the Central Limit Theorem and subtraction of normal variables.
Central Limit Theorem
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.
For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean [tex]\mu = p[/tex] and standard deviation [tex]s = \sqrt{\frac{p(1-p)}{n}}[/tex]
Subtraction of normal variables:
When we subtract normal variables, the mean is the subtraction of the means, while the standard error is the square root of the sum of the variances:
Of the 193 smokers taking a placebo, 29 stopped smoking by the 8th day.
This means that:
[tex]p_S = \frac{29}{193} = 0.1503[/tex]
[tex]s_S = \sqrt{\frac{0.1503*0.8497}{193}} = 0.0257[/tex]
Of the 266 smokers taking only the antidepressant buproprion, 82 stopped smoking by the 8th day.
This means that:
[tex]p_A = \frac{82}{266} = 0.3083[/tex]
[tex]s_A = \sqrt{\frac{0.3083*0.6917}{266}} = 0.0283[/tex]
Calculate the estimated standard error for the sampling distribution of differences in sample proportions.
[tex]s = \sqrt{s_S^2 + s_A^2} = \sqrt{0.0257^2 + 0.0283^2} = 0.0382[/tex]
The estimated standard error for the sampling distribution of differences in sample proportions is 0.0382.
